# Does this group action fix a point?

I saw a question on a forum earlier, and I'm curious to see how it can be solved.

Suppose that $G$ is a group of order $pr$ for distinct primes $p$ and $r$, and let $G$ act on a set $S$ of order $pr-p-r$.

I curious to see if there is a point $s\in S$ such that $gs=s$ for all $g\in G$. One can start off by supposing no such point exists. Then the stabilizer for each $s\in S$ is not of order $pr$, so there is no orbit of order $1$. Thus every orbit of any $s\in S$ must have order $p$, $r$, or $pr$. However, an orbit of $pr$ would be too large, so each possible orbit has order $p$ or $r$. Is there some way to derive a contradiction based on this to get the result? If not, how can it be done?

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Your missing piece is in this question: If $p$ and $r$ are coprime (thus especially when they are distinct primes), then $pr-p-r$ is the largest number that is not a sum of non-negative multiples of $p$ and $r$. So there is no way to partition $S$ into orbits of the specified sizes.